[Php Form [Solved]]

Fantastic, that done the trick.  :)

[Php Form [Solved]]

Its seems to be working.

[code]$_GET: Array
(
)
$_POST: Array
(
    [select] => 2
)
[/code]

Is what it returns, but surely that would mean that the $_POST should have worked?

[Php Form [Solved]]

It returns the following
[quote]
Warning: Wrong parameter count for print_r() in \\www\ViewDetails.php on line 16

$_GET:


Warning: Wrong parameter count for print_r() in \\www\ViewDetails.php.php on line 17

$_POST: [/quote]

Do I have to give a parameter for that to work?

[Php Form [Solved]]

Yehh I tryed that one as well, but it doesnt get a value either.

[Php Form [Solved]]

Right I've got a problem linking from this.

I have set up the next page which which has the  [code]$ID=$HTTP_POST_VARS['ID'];[/code]

To get the value from the above, the only problem is it doesnt get the value.

Any Ideas?

[Php Form [Solved]]

LOL, it always seems so obvious after someone points it out to you!

Thanks guys!

[Php Form [Solved]]

I'm trying to get a drop down menu to work, but I can't seem to get it to work at all. Its got to be in the <?php ?> code as its to get values dynamically.

The list is going to be filled with 2 fields from a MySQL database. The Name field is going to be the text shown in the drop down menu, the value is going to be the value from ID colum in the database.

[code]
<?php
$result = mysql_query("SELECT * FROM client")
or die(mysql_error()); 



echo '<form name="form1" method="post" action="">';
echo '<select name="select">';
//make some short variables
$ID=$row['ID'];//suppy the correct field names below
$Name=$row['Name'];
while($row = mysql_fetch_array( $result )) {
echo '<option value="' . $ID . '">' . $Name . '</option>';
}
echo "</select>";
echo "</form>";

?>[/code]

Although the menu is filled with the correct number of blank entries, but the every option is a " and every value is ' ' Does anyone know what I'm doing wrong?